FS6

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Find the Fourier series for f(x)=x^{2}\, on [-\pi ,\pi ]\,

The general Fourier series on [-L,L]\, is:

f(x)={\frac  {a_{0}}{2}}+\sum _{{k=1}}^{\infty }a_{k}\,\cos({\frac  {k\pi x}{L}})+b_{k}\,\sin({\frac  {k\pi x}{L}})\,

a_{n}={\frac  {1}{L}}\int _{{-L}}^{L}f(x)\cos({\frac  {n\pi x}{L}})\,dx,\,\,\,n=0,1,2,...\,

b_{n}={\frac  {1}{L}}\int _{{-L}}^{L}f(x)\sin({\frac  {n\pi x}{L}})\,dx,\,\,\,n=1,2,3,...\,

In the present problem, L=\pi \,

a_{k}={\frac  {1}{\pi }}\int _{{-\pi }}^{\pi }x^{2}\cos(kx)\,dx={\frac  {4(-1)^{k}}{k^{2}}}\,\,\,k=1,2,3,...\,

a_{0}={\frac  {1}{\pi }}\int _{{-\pi }}^{\pi }x^{2}\,dx={\frac  {1}{3\pi }}[x^{3}]_{{-\pi }}^{\pi }={\frac  {2\pi ^{2}}{3}}\,

b_{k}={\frac  {1}{\pi }}\int _{{-\pi }}^{\pi }x^{2}\sin(kx)\,dx=0\,\,\,k=1,2,3,...\,

So the Fourier series is:

x^{2}={\frac  {\pi ^{2}}{3}}+\sum _{{k=1}}^{\infty }{\frac  {4(-1)^{k}}{k^{2}}}\cos(kx)\, on [-\pi ,\pi ]\,

Fourier Series

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